You have found the following ages (in years) of all 6 zebras at your local zoo: $ 11,\enspace 5,\enspace 1,\enspace 15,\enspace 17,\enspace 16$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{11 + 5 + 1 + 15 + 17 + 16}{{6}} = {10.8\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $11$ years $0.2$ years $0.04$ years $^2$ $5$ years $-5.8$ years $33.64$ years $^2$ $1$ year $-9.8$ years $96.04$ years $^2$ $15$ years $4.2$ years $17.64$ years $^2$ $17$ years $6.2$ years $38.44$ years $^2$ $16$ years $5.2$ years $27.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.04} + {33.64} + {96.04} + {17.64} + {38.44} + {27.04}} {{6}} $ $ {\sigma^2} = \dfrac{{212.84}}{{6}} = {35.47\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{35.47\text{ years}^2}} = {6\text{ years}} $ The average zebra at the zoo is 10.8 years old. There is a standard deviation of 6 years.